Problem: Is ${569217}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {569217}= &&{5}\cdot100000+ \\&&{6}\cdot10000+ \\&&{9}\cdot1000+ \\&&{2}\cdot100+ \\&&{1}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {569217}= &&{5}(99999+1)+ \\&&{6}(9999+1)+ \\&&{9}(999+1)+ \\&&{2}(99+1)+ \\&&{1}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {569217}= &&\gray{5\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {5}+{6}+{9}+{2}+{1}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${569217}$ is divisible by $3$ if ${ 5}+{6}+{9}+{2}+{1}+{7}$ is divisible by $3$ Add the digits of ${569217}$ $ {5}+{6}+{9}+{2}+{1}+{7} = {30} $ If ${30}$ is divisible by $3$ , then ${569217}$ must also be divisible by $3$ ${30}$ is divisible by $3$, therefore ${569217}$ must also be divisible by $3$.